Problem: $\dfrac{ -4c + d }{ 5 } = \dfrac{ -2c - 9e }{ 2 }$ Solve for $c$.
Solution: Multiply both sides by the left denominator. $\dfrac{ -4c + d }{ {5} } = \dfrac{ -2c - 9e }{ 2 }$ ${5} \cdot \dfrac{ -4c + d }{ {5} } = {5} \cdot \dfrac{ -2c - 9e }{ 2 }$ $-4c + d = {5} \cdot \dfrac { -2c - 9e }{ 2 }$ Multiply both sides by the right denominator. $-4c + d = 5 \cdot \dfrac{ -2c - 9e }{ {2} }$ ${2} \cdot \left( -4c + d \right) = {2} \cdot 5 \cdot \dfrac{ -2c - 9e }{ {2} }$ ${2} \cdot \left( -4c + d \right) = 5 \cdot \left( -2c - 9e \right)$ Distribute both sides ${2} \cdot \left( -4c + d \right) = {5} \cdot \left( -2c - 9e \right)$ $-{8}c + {2}d = -{10}c - {45}e$ Combine $c$ terms on the left. $-{8c} + 2d = -{10c} - 45e$ ${2c} + 2d = -45e$ Move the $d$ term to the right. $2c + {2d} = -45e$ $2c = -45e - {2d}$ Isolate $c$ by dividing both sides by its coefficient. ${2}c = -45e - 2d$ $c = \dfrac{ -45e - 2d }{ {2} }$